Prime Numbers & Functions: Problem Solving In Math
Hey guys, ever get those math problems that seem like they're written in another language? Well, let's break down some tricky ones together! We're going to tackle problems involving prime numbers, functions, and a bit of number theory. So, grab your thinking caps, and let's dive in!
Understanding the Function: f(n) = n² - 79n + 1601
Our first problem revolves around a function, f(n) = n² - 79n + 1601. This looks a bit intimidating, but we'll take it one step at a time. The core idea here is to understand how this function behaves and what it tells us about numbers, particularly prime numbers. Prime numbers, as you know, are whole numbers greater than 1 that are only divisible by 1 and themselves (like 2, 3, 5, 7, etc.). We're going to explore how this function interacts with these special numbers.
(a) Is f(25) + f(30) a Prime Number?
Let's start with the first part: determining whether f(25) + f(30) is a prime number. To do this, we need to calculate the values of f(25) and f(30) using the function definition. First, let's calculate f(25):
f(25) = (25)² - 79(25) + 1601
f(25) = 625 - 1975 + 1601
f(25) = 251
Now, let's calculate f(30):
f(30) = (30)² - 79(30) + 1601
f(30) = 900 - 2370 + 1601
f(30) = 131
So, we have f(25) = 251 and f(30) = 131. Now we need to add these together:
f(25) + f(30) = 251 + 131 = 382
Finally, we need to determine if 382 is a prime number. A quick check reveals that 382 is divisible by 2 (382 = 2 * 191), so it is not a prime number. Therefore, the answer to the first part is no, f(25) + f(30) is not prime. This illustrates how evaluating a function at specific points can lead us to interesting questions about the nature of numbers.
(b) Proving f(n) = f(m) When m + n = 79
The second part of the problem presents us with a conditional statement: if m + n = 79, then we need to show that f(n) = f(m). This is where the algebraic properties of the function come into play. To prove this, we can start by expressing m in terms of n (or vice versa) using the given condition. Since m + n = 79, we can write m = 79 - n. Now, let's substitute this expression for m into the function f(m):
f(m) = f(79 - n) = (79 - n)² - 79(79 - n) + 1601
Now, we need to expand and simplify this expression:
f(79 - n) = (79² - 2 * 79 * n + n²) - (79² - 79n) + 1601
f(79 - n) = 6241 - 158n + n² - 6241 + 79n + 1601
f(79 - n) = n² - 79n + 1601
Notice anything familiar? The simplified expression for f(79 - n) is exactly the same as our original function, f(n). This means that f(m) = f(79 - n) = f(n), which is precisely what we needed to prove. This part highlights the symmetry inherent in the function and how specific conditions can lead to surprising equalities.
(c) Finding the Smallest Prime f(n)
The final part asks us to determine the value of n for which f(n) is the smallest prime number. This requires a bit of trial and error, combined with our understanding of the function's behavior. We already know that f(n) = n² - 79n + 1601. Let's start by trying some values of n and see what we get:
- For n = 0, f(0) = 1601 (which is 29 * 55.2, so not prime)
- For n = 1, f(1) = 1² - 79(1) + 1601 = 1523 (not easily divisible, let's keep this in mind)
- For n = 2, f(2) = 2² - 79(2) + 1601 = 1447 (not easily divisible either)
We can continue this process, but it might be helpful to consider the symmetry we discovered in part (b). We know that f(n) = f(79 - n). This means that the function will have the same value for n and 79 - n. For example, f(1) will be the same as f(78), f(2) will be the same as f(77), and so on. This symmetry suggests that the smallest value of f(n) might occur somewhere in the middle, around n = 39 or 40.
Let's try n = 39:
f(39) = (39)² - 79(39) + 1601
f(39) = 1521 - 3081 + 1601
f(39) = 41
And let's check n = 40:
f(40) = (40)² - 79(40) + 1601
f(40) = 1600 - 3160 + 1601
f(40) = 41
We found that f(39) = f(40) = 41. Since 41 is a prime number, and we've explored values around the minimum due to the symmetry, it's highly likely that 41 is the smallest prime value of f(n). To be absolutely sure, we could check values closer to 0 and 79, but given the quadratic nature of the function, the values will increase as we move away from the center. Thus, the value of n for which f(n) is the smallest prime is either 39 or 40.
Divisibility Proof: 3(2n) + 1
Now, let's move on to the second problem. We need to prove that 3(2n) + 1 is divisible by a prime number if n is a positive integer. This problem ventures into the realm of number theory, specifically dealing with divisibility and exponents. To tackle this, we can think about how numbers of this form behave and look for potential patterns or factors.
Exploring the Expression
Let's start by looking at the first few values of the expression for different values of n:
- For n = 1, 3(21) + 1 = 3² + 1 = 9 + 1 = 10 (divisible by 2 and 5)
- For n = 2, 3(22) + 1 = 3â´ + 1 = 81 + 1 = 82 (divisible by 2 and 41)
- For n = 3, 3(23) + 1 = 3⸠+ 1 = 6561 + 1 = 6562 (divisible by 2 and 3281)
From these examples, we can see that the result is always an even number (divisible by 2). This is a crucial observation! We can prove this generally.
Proving Divisibility by 2
To prove that 3(2n) + 1 is always divisible by 2, we need to show that it's an even number. An even number can be expressed in the form 2k, where k is an integer. Let's analyze the expression:
- 3 raised to any positive integer power will always be an odd number. This is because 3 is odd, and the product of odd numbers is always odd.
- Therefore, 3(2n) is odd.
- Adding 1 to an odd number always results in an even number.
- Thus, 3(2n) + 1 is even.
Since 3(2n) + 1 is even, it is divisible by 2. This completes the proof that 3(2n) + 1 is divisible by a prime number (specifically, 2) when n is a positive integer.
Conclusion
So, guys, we've journeyed through some interesting math problems today! We explored the behavior of a quadratic function, delved into the world of prime numbers, and even proved a divisibility rule. These kinds of problems might seem tough at first, but breaking them down step-by-step and looking for patterns can make them much more manageable. Keep practicing, keep exploring, and you'll be amazed at what you can discover in the world of math!
